Is $ \frac{2}{1+e^{t^2}} $ a characteristic function?

Question

I'm trying to establish whether the following is a characteristic function of some random variable:

$$ \phi(t) = \frac{2}{1+ e^{t^2}} .$$

It satisfies all basic characteristic function properties, being one at zero, uniformly continuous, and since it's a real-valued function, if it indeed is a characteristic function, then of some symmetric distribution.

I don't know how to go on.

Answer 1

Let's see that

$$ \frac{2}{1+ e^{t^2}} (\frac{1}{2} + \frac{1}{2e^{t^2}}) = e^{-t^2} $$

and the second part of the product is a characteristic function (since it is a convex combination of such). This equality contradict Cramer's theorem which states that if $ Z = X + Y $ and $ X,Y $ are independent, $ Z $ is normally distributed, then $ X,Y $ are normally distributed